Kinematics

MDCAT Physics · Physics · Force and motion. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

Kinematics describes motion without asking what causes it. Displacement is the straight-line change of position with its direction, so it can be smaller than the distance travelled. Velocity is the rate of change of displacement, and acceleration is the rate of change of velocity.

When the acceleration is uniform, four equations link u, v, a, s and t. Choose one direction as positive and keep to it. In free fall the acceleration is g downwards at every point, even at the top of the path where the velocity is zero.

Graphs carry the same information. The slope of a displacement–time graph is the velocity. The slope of a velocity–time graph is the acceleration, and the area under it is the displacement.

Rules to remember

average velocity = total displacement ÷ total time; average speed = total distance ÷ total time
v = u + at
s = ut + ½at² and s = ½(u + v)t
v² − u² = 2as
From rest with uniform acceleration, distances in the 1st, 2nd, 3rd, 4th seconds are in the ratio 1 : 3 : 5 : 7
Displacement–time graph: slope = velocity. Velocity–time graph: slope = acceleration, area = displacement
Area below the time axis counts as negative for displacement but is added for distance

Common mistake

Students say the acceleration of a ball is zero at the top of its flight. Only the velocity is zero there; the acceleration is still g downwards.

Worked example

A train moving at 30 m s⁻¹ brakes uniformly and stops after travelling 225 m. Find its deceleration and the time it takes to stop.

Use v² − u² = 2as: 0 − 30² = 2 × a × 225, so a = −900/450 = −2.0 m s⁻².
Use v = u + at: 0 = 30 − 2.0t, so t = 15 s.
Check with s = ½(u + v)t = ½ × 30 × 15 = 225 m.

Answer: Deceleration 2.0 m s⁻²; time 15 s.

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